Rudin's Real and Complex Analysis -upto Section 1.9 review:
At this point (if you are following this from previous post, Rudin introduces a Theorem $1.7$ that is fundamental and very useful.
Stated informally,
1. Continuous function of continuous function is continuous.
2. Continuous function of measurable function is measurable.
(2) requires $Y,Z$ are topological spaces for continuous functions and $X$ is measurable. Of course (1) requires all spaces be topological.
The proof relies on definitions of continuity and measurability. Measurability requires $f^{-1}(V)$ of any open set $V$ be measurable.
Theorem 1.8 is very puzzling at first. There is no explanation why we need a plane and what it is doing here. Fortunately, reading next section ($1.9$) makes it clear. Also it uses a fact from real analysis - every open set in plane is a countable union of rectangles with sides parallel to axis.
The proof is somewhat complicated - but not too bad. Given $u(x),v(x)$ are measurable, it starts off by setting $f=(u(x),v(x))$ as mapping of $X$ (measurable space) into a plane and setting $h=\Phi\circ f$ and thus requiring measurability of $f$ to complete the proof. Since $\Phi$ is defined to be continuous function, the measurability of $f$ completes the proof. To show $f$ is measurable all that's needed to show is $f^{-1}(V)$ for any open set $V$ should be measurable.
At this point (if you are following this from previous post, Rudin introduces a Theorem $1.7$ that is fundamental and very useful.
Stated informally,
1. Continuous function of continuous function is continuous.
2. Continuous function of measurable function is measurable.
(2) requires $Y,Z$ are topological spaces for continuous functions and $X$ is measurable. Of course (1) requires all spaces be topological.
The proof relies on definitions of continuity and measurability. Measurability requires $f^{-1}(V)$ of any open set $V$ be measurable.
Theorem 1.8 is very puzzling at first. There is no explanation why we need a plane and what it is doing here. Fortunately, reading next section ($1.9$) makes it clear. Also it uses a fact from real analysis - every open set in plane is a countable union of rectangles with sides parallel to axis.
The proof is somewhat complicated - but not too bad. Given $u(x),v(x)$ are measurable, it starts off by setting $f=(u(x),v(x))$ as mapping of $X$ (measurable space) into a plane and setting $h=\Phi\circ f$ and thus requiring measurability of $f$ to complete the proof. Since $\Phi$ is defined to be continuous function, the measurability of $f$ completes the proof. To show $f$ is measurable all that's needed to show is $f^{-1}(V)$ for any open set $V$ should be measurable.
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